Expectation of an integral of a function of a Brownian motion

Question

$B_t$ is a Brownian motion and $Y_t:=e^{aB_t+bt}$. For which $a$ and $b$ is $Y_t\in M^2$?

I found a theorem that says that sufficient for $Y_t\in M^2$ would be $E[\int_0^\infty Y_t^2 dt]<\infty $

But how can I integrate over a function of a Brownian motion?

I don't think that the Ito isometry is helpful, because I would need $Y_t\in M^2$ before I can use it.

Or is there a simpler way to prove $Y_t\in M^2$ or $E[\int_0^\infty Y_t^2 dt]<\infty $ without having to calculate the integral?

Answer 1

Evaulte

$$E\left[\int_0^\infty Y_t^2 dt\right] = E\left[\int_0^\infty e^{2aB_t+2bt} dt\right] $$

$$= \int_0^\infty E\left[e^{2aB_t}\right] e^{2bt} dt=\int_0^\infty e^{2a^2t}e^{2bt} dt $$ $$=\frac{1}{2(a^2+b)}e^{2(a^2+b)t}|_0^\infty $$

Thus, the condition on $a$ and $b$ is

$$a^2+b < 0$$

Answer 2

• If $$\mathbb E\int_0^\infty Y_t^2dt<\infty $$ then $\left(\int_0^t Y_sdB_s\right)_{t\geq 0}$ is a martingale, but in general, not $(Y_t)_{t\geq 0}$.

• Let $s<t$. $$\mathbb E[e^{aB_t+bt}\mid \mathcal F_s]=e^{bt+aB_s}\mathbb E[e^{a(B_t-B_s)}\mid \mathcal F_s]=e^{bt+aB_s}\mathbb E[e^{a(B_t-B_s)}].$$ Using the fact that $B_t-B_s\sim \mathcal N(0,t-s)$ allow you to find $a,b$ s.t. $(e^{aB_t+bt})_{t\geq 0}$ is a Martingale. Now, $\mathbb E[Y_t^2]$ shouldn't be to complicate to calculate, and will allow you to conclude.

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An other way is using Itô formula, $$e^{aB_t+bt}=1+\int_0^t\left(\frac{a^2}{2}+b\right)e^{aB_t+bt}\,\mathrm d t+\int_0^tae^{aB_s+bs}\,\mathrm d B_s.$$

As far as you proved that $$\mathbb E\int_0^\infty e^{2aB_t+2bt}\,\mathrm d t<\infty ,$$ then, $(e^{aB_t+bt})_{t\geq 0}$ is a Martingale $\iff$ $\frac{a^2}{2}+b=0$.